\(n_{HCl}=\dfrac{250.18,25\%}{100\%}:36,5=1,25\left(mol\right)\)
\(n_{H_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_A=x\left(mol\right)\\n_B=x\left(mol\right)\end{matrix}\right.\) (theo đề)
\(A+2HCl\rightarrow ACl_2+H_2\)
x---->2x---------------->x
\(2B+6HCl\rightarrow2BCl_3+3H_2\)
x----->3x------------------->1,5x
Theo PTHH có: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.1,25=0,625\left(mol\right)>n_{H_2.thoát.ra.theo.đề}\)
\(\Rightarrow HCl.dư\\ \Rightarrow n_{H_2}=x+1,5x=0,375\\ \Rightarrow x=0,15\)
Có: \(m_{hh.kim.loại}=m_A+m_B=Ax+Bx=x\left(A+B\right)=7,65\) (g)
\(\Rightarrow A+B=\dfrac{7,65}{0,15}=51\left(g/mol\right)\)
Mà A là kim loại hóa trị II, B là kim loại hóa trị III.
\(\Rightarrow\left\{{}\begin{matrix}A=24\left(Mg\right)\\B=27\left(Al\right)\end{matrix}\right.\)